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We will use the notation card(A) for the cardinal number of a set A.

The conditions for Ω are as follows:

(1)φ≠Ω;

(2)for all A in Ω, card(A)≤card(N) or card(A^c)≤card(N).

On the other hand, Ω is a σ-field if and only if

<1>φ∈Ω;

<2>A∈Ω ⇒ A^c∈Ω;

<3>countable union and countable intersection are in Ω.

We will show that Ω in question satisfies <1> through <3>.

<1> trivially holds.

<2> According to the condition (2), if A is in Ω, then

card(A)≤card(N) or card(A^c)≤card(N). Whence,

card(A^c)≤card(N) or card(A)=card((A^c)^c) ≤card(N), which show A^c∈Ω.

<3>First of all, we will prove

A,B∈Ω ⇒ A⋂B∈Ω

A,B∈Ω ⇒ A⋃B∈Ω.

The former:

When A,B∈Ω, the following 4 cases arise:

card(A)≤card(N) and card(B)≤card(N);

card(A)≤card(N) and card(B^c)≤card(N);

card(A^c)≤card(N) and card(B)≤card(N);

card(A^c)≤card(N) and card(B^c)≤card(N).

If (a) happens,

card(A⋂B)≤card(A) and card(B)≤card(N).

If (b) happens,

card(A⋂B)≤card(A) and card(B)≤card(N).

If (c) happens,

card(A⋂B)≤card(A) and card(B)≤card(N).

If (d) happens,

card((A⋂B)^c)=card((A^c)⋃(B^c))≤card(A^c)+card(B^c)

≤card(N)+card(N)=card(N).

Thus, we have A⋂B∈Ω.

The latter:

A⋃B=((A^c)⋂(B^c))^c. But A^c, B^c are both in Ω(by <2>,

so (A^c)⋂(B^c) is in Ω(by the preceding result), and

((A^c)⋂(B^c))^c is in Ω(by <2>).

Hence, A⋃B is in Ω.

*General Countable Cases*

Let An (n=1,2,…) be countable elements in Ω.

We can observe that ⋂_(n=1)^k▒An=(⋂_(n=1)^(k-1)▒An)⋂Ak.

So, closure of countable intersection follows by induction.

And also closure of countable union follows.

Therefore, Ω is aσ-field.

I confess that I don't know where to begin, but at the beginning. We need to show 3 things:

- set is not empty

- set is closed under complement

- set is closed under countable union of sets

Showing that the set is not empty should be easy because even if subset A were empty, it's complement would not be.

Showing the set is closed under complement could be tricky, but it almost seems as though if it is closed under complementary operation, then a countable union of the subsets (or their complements) would be within the set as well.

What is this: Lebesgue integration? A bit highbrow for me.